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Question

A Norman window has the shape of a rectangle surmounted by a semicircle. A norman window with perimeter 30 ft is to be constructed.

Find a function that models the window.
Find the dimensions of the window that admits the greatesr amount of light.

The base of the window is x

Answer 1

perimeter =2 r+2h +pi*r =30 so h= [30-(2+pi)r]/2 and the surface
S=30r-(2+pi)r^2+pir^2/2
dS/dr=[30-2(2+pi)r +2pir]=0
30-4r=0 r=7.5 feet
So the base is 15 feet and the height of the rectangle=
12.43 feet
correction
15+2h +23.56=30 so h is negative (impossible)

Answer 2

So x is the width of the rectangle and let y be the length of the rectangle. the width of the rectangle is also the radius of the semicircle.

the perimeter of the window is:
P = .5 (2 π r) + y + y + x

given that the perimeter is 30ft

30 = π r + 2y + x
30 = π x + 2y + x

the area of the window is the sum of the area of the rectangle and the area of the semicircle

A = .5π r² + xy
A = .5π x² + xy

30 = π x + 2y + x

solve for y
30 - πx - x = 2y
y = 15 - πx/2 - x/2

substitute
A = .5π x² + xy
A = .5π x² + x(15 - πx/2 - x/2)
A = .5π x² + 15x - πx²/2 - x²/2
A = -x²/2 + 15x

now graph the equation and find the maximum point.
the maximum area is the y value of the vertex. The base of the rectangle is the x value of the vertex

vertex: (15,112.5)

so x = 15ft